Factor polynomial $$ 15x^{2}+14x-8 $$

$$ 15x^{2}+14x-8 = ( 5x-2 )( 3x+4 ) $$

Step 1: Identify constants $ a $ , $ b $ and $ c $.
$ a $ is a number in front of the $ x^2 $ term $ b $ is a number in front of the $ x $ term and $ c $ is a constant. In this case:

$$ a = 15 , b = 14 ~ \text{ and } ~ c = -8 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 15 }$ by the constant term $\color{blue}{c = -8} $.

$$ a \cdot c = -120 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -120 $ and add to $ b = 14 $.

Step 4: All pairs of numbers with a product of $ -120 $ are:

PRODUCT = -120
-1     120	1     -120
-2     60	2     -60
-3     40	3     -40
-4     30	4     -30
-5     24	5     -24
-6     20	6     -20
-8     15	8     -15
-10     12	10     -12

Step 5: Find out which factor pair sums up to $\color{blue}{ b = 14 }$

PRODUCT = -120 and SUM = 14
-1     120	1     -120
-2     60	2     -60
-3     40	3     -40
-4     30	4     -30
-5     24	5     -24
-6     20	6     -20
-8     15	8     -15
-10     12	10     -12

Step 6: Replace middle term $ 14 x $ with $ 20x-6x $:

$$ 15x^{2}+14x-8 = 15x^{2}+20x-6x-8 $$

Step 7: Apply factoring by grouping. Factor $ 5x $ out of the first two terms and $ -2 $ out of the last two terms.

$$ 15x^{2}+20x-6x-8 = 5x\left(3x+4\right) -2\left(3x+4\right) = \left(5x-2\right) \left(3x+4\right) $$

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